PROFESSOR: Welcome to this session
on separable equations.
So in this problem, you're asked in the first question to
solve the initial value problem, dydx equals y squared
with the initial condition y of zero equals 1.
In the second part of the problem, you're asked to find
the general solution where no initial condition is imposed.
So here you need to remember your method of separation of
variables to tackle the first question.
And then in the second part of the problem, remember all the
types of solutions and conditions that you applied
the first part to recover the lost solutions.
So why don't you take a minute, pause the video, work
through questions a and b.
And then we'll continue together when I come back.
Welcome back.
So in the first part of the problem, we'll be solving the
equation dydx equals y squared.
So here the method of separation of variables tells
us that we should regroup the variables of the same kind, so
all the y variables on one side and dx variable on the
other side of the equation and then
integrate from this point.
So here for this step, notice that I divided by y squared,
which means that we need to impose the condition y naught
equal to zero from now on.
So from this step, we just use indefinite integrals to
integrate both sides of the equation.
So the left-hand side is the integral dy over y squared.
So integral of this gives us minus 1 over y.
And the right hand side, integral dx, is just x.
Both sides would give us constant of integrations, but
we only need one because this is a first-order
differential equation.
And so we group them together on the right-hand side with
constant c.
So from this point, given that we're interested in variable
y, we need just to invert the expression.
And that gives us partial answer, which is y of x equals
minus 1 over x plus c.
So now we need to use our initial condition, y of zero
equals to 1 to determine the value of c for this particular
initial value problem.
So our initial condition was y of zero equals 1.
So if we substitute this in the expression that we just
obtained, we just have zero plus c, which then only gives
us c equal to 1.
So we end up with the value for our constant of
integration, c equals minus 1.
And so the solution to this problem is y of x equals 1
over 1 minus x.
So if you examine this expression, you see right away
that we have a problem for x equals 1 because at x equals
1, we have 1/0 which means that, then, the solution blows
up and we have a vertical asymptote.
So let me draw this here.
So we're going to have an asymptote at x equals 1 and a
solution that passes through our initial condition, y
equals 1, going to infinity when approaching x equals 1.
But then on the right side of the value x equals 1, we have
another part of the solution that goes to zero
as x goes to infinity.
And that diverges to minus infinity when x approaches 1.
So by convention, the solutions of differential
equations are defined on one single interval.
So we need here to realize that the solution we had is
basically two parts, the parts on the left of the asymptote
and the part on the right of the asymptote.
So the solution to our initial value problems needs to be the
solution that passes through the imposed initial condition,
which was y of zero equals to 1.
So it needs to be this solution.
So now, if we move on to the solution of the second part of
the problem b, we were asked to find the general solution
of the problem, which means that we need to account now
for all the solutions, regardless of
their initial condition.
So we already answered this partially during the solution
of part a where we solved using indefinite integrals and
arrived to the solution minus 1 over x plus c, where here we
had, basically, an undetermined constant of
integration.
So this is one general solution.
But remember that we need to give all the
solutions of the problem.
So when we arrived at the solution, we excluded the
solution y equals zero, which was basically a lost solution
because we had to impose the condition y not equal to zero.
So we need when we give the general solution to this
differential equation to recover the lost solution.
And then we basically have one kind of solution, minus 1 over
x plus c, that excludes y equal to zero and another kind
of solution that is simply the zero solution.
So to summarize, the important points of this problem is to
remember the separation of variables and how to use it
and the fact that using it imposes conditions that
require us to recover lost solutions at the end of the
problem if we're asked to give general solution.
Another point to remember is that even a simple ODE can
lead to relatively complex behavior which drives the
presence of this vertical asymptote that you need to
then know how to deal with and determine which part of the
solutions that you've obtained is the real solution to the
initial value problem that you're given.
So I hope that you are OK with this problem and you will use
these approaches many times for the rest of the course.